Water is leaking out of an inverted conical tank at a rate of 5800 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 8 meters and the diameter at the top is 6.5 meters. If the water level is rising at a rate of 28 centimeters per minute when the height of the water is 1 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
To find the rate at which water is being pumped into the tank, we can use related rates. Let's denote the rate at which water is being pumped into the tank as V (in cubic centimeters per minute).
We are given that the water level is rising at a rate of 28 centimeters per minute when the height of the water is 1 meter. This can be written as dh/dt = 28 cm/min, where h is the height of the water in meters and t is the time in minutes.
We are also given that the water is leaking out of the tank at a rate of 5800 cubic centimeters per minute. This can be written as dV/dt = -5800 cm^3/min, where V is the volume of the water in the tank.
To solve for dV/dt, we need to relate the volume of the water to its height. The tank is in the shape of an inverted cone, so we can use the formula for the volume of a cone V = (1/3)πr^2h, where r is the radius of the top of the cone.
Given that the diameter at the top of the cone is 6.5 meters, we can find the radius by dividing the diameter by 2: r = 6.5/2 = 3.25 meters.
To relate the height of the water to the radius, we can use similar triangles. Since the cone has a height of 8 meters, the ratio of the height of the water h to the height of the cone 8 is equal to the ratio of the radius of the water to the radius of the cone, r. Thus, h/8 = r/3.25.
Simplifying this equation, we get h = (8/3.25)r.
Now, we can substitute this expression for h into the volume formula to get V = (1/3)πr^2(8/3.25)r.
Differentiating both sides of this equation with respect to t, we get dV/dt = (1/3)π(8/3.25)(2r(dr/dt)).
Now, we need to find the value of dr/dt, which represents the rate at which the radius is changing with respect to time. Since the radius is half the diameter, which remains constant, dr/dt = 0.
Substituting this value into the equation for dV/dt, we get dV/dt = (1/3)π(8/3.25)(2r(0)) = 0.
Therefore, the rate at which water is being pumped into the tank is 0 cubic centimeters per minute.